## Volume formula derivations::Volume

### ::concepts

Volume::volume    Center::style    Radius::geometry    Sphere::''h''    Height::cubic    Manifold::integral

Volume formula derivations

### Sphere

The volume of a sphere is the integral of an infinite number of infinitesimally small circular disks of thickness dx. The calculation for the volume of a sphere with center 0 and radius r is as follows.

The surface area of the circular disk is $\pi r^2$.

The radius of the circular disks, defined such that the x-axis cuts perpendicularly through them, is

$y = \sqrt{r^2-x^2}$

or

$z = \sqrt{r^2-x^2}$

where y or z can be taken to represent the radius of a disk at a particular x value.

Using y as the disk radius, the volume of the sphere can be calculated as $\int_{-r}^r \pi y^2 \,dx = \int_{-r}^r \pi(r^2 - x^2) \,dx.$

Now $\int_{-r}^r \pi r^2\,dx - \int_{-r}^r \pi x^2\,dx = \pi (r^3 + r^3) - \frac{\pi}{3}(r^3 + r^3) = 2\pi r^3 - \frac{2\pi r^3}{3}.$

Combining yields $V = \frac{4}{3}\pi r^3.$

This formula can be derived more quickly using the formula for the sphere's surface area, which is $4\pi r^2$. The volume of the sphere consists of layers of infinitesimally thin spherical shells, and the sphere volume is equal to

$\int_0^r 4\pi u^2 \,du$ = $\frac{4}{3}\pi r^3.$

### Cone

The cone is a type of pyramidal shape. The fundamental equation for pyramids, one-third times base times altitude, applies to cones as well.

However, using calculus, the volume of a cone is the integral of an infinite number of infinitesimally thin circular disks of thickness dx. The calculation for the volume of a cone of height h, whose base is centered at (0,0,0) with radius r, is as follows.

The radius of each circular disk is r if x = 0 and 0 if x = h, and varying linearly in between—that is, $r\frac{(h-x)}{h}.$

The surface area of the circular disk is then $\pi \left(r\frac{(h-x)}{h}\right)^2 = \pi r^2\frac{(h-x)^2}{h^2}.$

The volume of the cone can then be calculated as $\int_{0}^h \pi r^2\frac{(h-x)^2}{h^2} dx,$

and after extraction of the constants: $\frac{\pi r^2}{h^2} \int_{0}^h (h-x)^2 dx$

Integrating gives us $\frac{\pi r^2}{h^2}\left(\frac{h^3}{3}\right) = \frac{1}{3}\pi r^2 h.$

Volume sections
Intro  Units   Related terms    Volume in calculus    Volume formulas    Volume formula derivations    Volume in differential geometry    Volume in thermodynamics    See also   References  External links

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